# Thread: forces problem

1. ## forces problem

OPQR is a horizontal plane rectangle of negigible mass. Its side lengths are OP =RQ =2m and OR = PQ = 1m. Forces act along each side of the rectangle.

If OP is taken to be i direction and j to be the direction of OR. The force K is given by -40i. If the rectangle is in equilibrium what is the force acting along PQ?

I think the answer is 20J- anyone help?

2. Originally Posted by thermalwarrior
OPQR is a horizontal plane rectangle of negigible mass. Its side lengths are OP =RQ =2m and OR = PQ = 1m. Forces act along each side of the rectangle.

If OP is taken to be i direction and j to be the direction of OR. The force K is given by -40i. If the rectangle is in equilibrium what is the force acting along PQ?

I think the answer is 20J- anyone help?
I'm not sure how to answer this. Where on the rectangle the -40 i N force applied? There obviously there must be a counterforce of 40 i N acting on the rectangle to hold it in place, but this force could be at various points of the rectangle or distributed say, along the whole side PQ. We need more information.

Also, the unit of force is N (Newtons), not J (Joules) which is a unit of energy.

-Dan

3. sorry j is that vertical axis (with i being the horizontal axis). -40i means there is 40N of force acting along the negative horizontal axis.

so you have a rectangle with two sides measuring 1m in the j axis and 2m in the i axis.

4. Originally Posted by thermalwarrior
sorry j is that vertical axis (with i being the horizontal axis). -40i means there is 40N of force acting along the negative horizontal axis.

so you have a rectangle with two sides measuring 1m in the j axis and 2m in the i axis.
I got that part. The question is, where is the -40 i N force applied? Is it applied to point O? R? Where it is applied will affect where the counterforce is to be applied. And for the force "acting along PQ": does that mean the force is distributed along the length of PQ or are we merely to say this is the force we have to apply at some point on PQ?

Taking this problem in the simplest sense, if we apply a force of -40 i N on the rectangle, then there must be a counterforce of 40 i N to keep it stationary. But as you are given the lengths of the sides, I doubt the problem is that simple and that you will have to take into account torques. Hence the need for more information.

-Dan

5. thanks for your reply - ive sorted it now taking the sum of the torques the answer was -20j.